Derive the K_(P) and K_(c) for the following equilibrium reaction. H_(2(g))+I_(2(g))hArr2HI_((g))

Derive the K_(P) and K_(c) for the following equilibrium reaction. H_

1.	Derive the K_(P) and K_(c) for the following equilibrium reaction. H_(2(g))+I_(2(g))hArr2HI_((g))
Answer» Solution :Let us consider the formation of HI in which, 'a' moles of hydrogen and 'b' moles of iodine gas are allowed to REACT in a container of valume V. Let 'x' moles of each of `H_(2)` and `I_(2)` react together to from 2X moles of HI. `H_(2(g))+I_(2(g))hArr2HI(g)` Applying law of mass action, `K_(c)=([HI]^(2))/([H_(2)][I_(2)])=(((2x)/(V))^(2))/(((a-x)/(V))((b-x)/(V)))=(4x^(2))/((a-x)(b-x))` The equilibrium CONSTANT `K_(P)` can also be calculated as follows : We KNOW the RELATIONSHIP between the `K_(c)` and `K_(P)` `K_(P)=K_(C)(RT)^((Deltan_(g))` Here the, `Deltan_((g))=n_(P)-n_(r)=2-2=0` Hence `K_(P)=K_(C),K_(P)=(4x^(2))/((a-x)(b-x))`